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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof</dfn> We have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
\begin{aligned}
&amp;y_1^{\prime \prime}+p(x) y_1^{\prime}+q(x) y_1=0,\\
&amp;y_2^{\prime \prime}+p(x) y_2^{\prime}+q(x) y_2=0.\\
\end{aligned}\tag{3.3.1}
\end{equation}
</div>
<p class="continuation">Besides, for the Wronskian, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;W(y_1, y_2)=\left|
\begin{array}{ll}
y_1 &amp; y_2\\
y_1^{\prime} &amp; y_2^{\prime}
\end{array}
\right|=y_1 y_2^{\prime}-y_2 y_1^{\prime},\\
&amp;\frac{\textrm{d}W(y_1, y_2)}{\textrm{d} x}=y_1 y_2^{\prime \prime}+y_1^{\prime} y_2 ^{\prime}-y_1^{\prime} y_2 ^{\prime}-y_2 y_1^{\prime \prime}=y_1 y_2^{\prime \prime}-y_2 y_1^{\prime \prime}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">From <span class="process-math">\((\ref{eq3_9})_2 \cdot y_1-(\ref{eq3_9})_1 \cdot y_2\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1 y_2^{\prime \prime}-y_2 y_1^{\prime \prime}+p(x) y_1 y_2^{\prime}-p(x) y_2 y_1^{\prime}=0~\rightarrow~\frac{\textrm{d} W}{\textrm{d}x}+p(x) W=0~\rightarrow~\frac{\textrm{d} W}{W}=-p(x) \textrm{d} x
\end{equation*}
</div>
<p class="continuation">which yields</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W=C \exp \left[-\int p(x) \textrm{d} x\right].
\end{equation*}
</div>
<span class="incontext"><a href="sec3_3.html#p-90" class="internal">in-context</a></span>
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